A rectangle has and area of 55 Square inches and a length of 5 inches what is the width

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago

If z varies inversely as x^2, and z = 9 when x = 2/3, find z when x = 5/4

zimovet [89]

Answer:

Step-by-step explanation:

z varies inversely

so z=a/x^2

9=a/(2/3)^2

9=a/(4/9)

9=9a/4

a=4

z=4/x^2

z=4/(5/4)^2

z=4/(25/16)

z=64/25

6 0

2 years ago

Can someone please help me

Dennis_Churaev [7]

Okay, so I had to resort to the calculator since I couldn't figure out how to do it, anyway here.

(-0.08928571428) -this is the answer according to it id says take a few numbers and see what answer fits it the most but yeah, I didn't help much.

3 0

1 year ago

Find the mean of the integers

mrs_skeptik [129]

We need to find mean of given integers.

We have -16,39,-10,-16,12,31.

In order to find mean, we need to add all the given inetgers.

Therefore, putting plus sign in between integers.

-16+ 39+ (-10)+(-16)+12+31 .

We will add all positive number 39+12+31 = 82

We will add all negative numbers by negative numbers, -16-10-16 = -42.

Therefore, -16+ 39+ (-10)+(-16)+12+31 = 82 - 42 = 40.

We have given total 6 integers.

So, in order to find the mean, we need to apply following formula

Mean = $\frac{Sum \ of \ all \ number }{Total \ numbers}$

We get,

Mean = 40/6 = 6.66666......

We can round it to 6.67.

Therefore, mean of the given integers = 6.67 (approximately) .

4 0

3 years ago

Can someone help me with this: y=-5x+5

NikAS [45]

Answer:

I think it would be y=x

Step-by-step explanation:

I might be wrong

8 0

2 years ago

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